Show the 3D Laplace equation has a solution…
Show the 3D Laplace equation has a solution $u=F(z+\sqrt{x^2+y^2+z^2})$ and find it.
Explanation of Method
Compute $\Delta\xi$ and $|\nabla\xi|^2$ for $\xi=z+r$ directly, then apply the chain rule $\Delta u=F''|\nabla\xi|^2+F'\Delta\xi$ to reduce $\Delta u=0$ to a simple ODE in $F$.
Step 1 · Laplacian and gradient-squared of $\xi=z+r$
With $r=\sqrt{x^2+y^2+z^2}$: direct differentiation gives $\Delta r=2/r$ and (after simplification using $\xi=z+r$) $\Delta\xi=2/r$ as well; similarly $|\nabla\xi|^2 = \frac{x^2+y^2+\xi^2}{r^2}$, and expanding $\xi^2=z^2+2zr+r^2$ shows $x^2+y^2+\xi^2=2r\xi$, so \[ |\nabla\xi|^2 = \frac{2\xi}{r}. \]
Step 2 · Reduce $\Delta u=0$ to an ODE
\[ \Delta u = F''(\xi)\cdot\frac{2\xi}{r}+F'(\xi)\cdot\frac2r = \frac2r\frac{d}{d\xi}\bigl[\xi F'(\xi)\bigr] = 0 \implies \xi F'(\xi)=C_1. \]
Step 3 · Integrate
\[ F'(\xi)=\frac{C_1}{\xi} \implies F(\xi)=C_1\ln\xi+C_2. \]
Verification
With $F'(\xi)=C_1/\xi$: $\xi F'(\xi)=C_1$ is constant, so its $\xi$-derivative is $0$, giving $\Delta u=0$. \checkmark
Geometric remark
The level surfaces $z+r=\text{const}$ are confocal paraboloids of revolution with focus at the origin, so this solution is the harmonic function analogous to $\ln r$ in 2D but with paraboloidal equipotentials instead of circular ones.
Result
The solution is: